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请教两道数学题

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楼主
发表于 2008-2-7 22:32:00 | 只看该作者

请教两道数学题

1.Of the families in City X in 1994, 40 percent owned a personal computer.  The number of families in City X owning a computer in 1998 was 30 percent greater than it was in 1994, and the total number of families in City X was 4 percent greater in 1998 than it was in 1994.  What percent of the families in City X owned a personal computer in 1998?

 

A.     50%

B.     52%

C.     56%

D.     70%

E.      74%

key is A. why?

2.Last Sunday a certain store sold copies of Newspaper A for $1.00 each and copies of Newspaper B for $1.25 each, and the store sold no other newspapers that day.  If r percent of the store’s revenues from newspaper sales was from Newspaper A and if p percent of the newspapers that the store sold were copies of newspaper A, which of the following expresses r in terms of p?

 

A.     100p / (125 – p)

B.     150p / (250 – p)

C.     300p / (375 – p)

D.     400p / (500 – p)

E.      500p / (625 – p)

key is D. why?

many thanks!!!

沙发
发表于 2008-2-8 15:50:00 | 只看该作者
1. 假设1994年City X有A户家庭,则有电脑的为 40%A 户家庭。

有题干可知,到1998年City X有电脑的为 40%A(1+30%)=52%A 户家庭;并且到1998年,City X有 (1+4%)A=104%A 户家庭。

由此可知,有电脑的家庭占总家庭数的比例为:52%A/104%A=50%

2. 总收入由TR表示,卖出报纸总数量由Q表示,卖出报纸A数量由A表示,卖出报纸B数量由B表示。

有题干可知,$1.00A+$1.25B=TR(即A+1.25B=TR),$1.00A=r%TR(即A=r%TR),A=p%Q;

由后两个等式可得出A=r%TR=p%Q,TR=(p/r)Q;由该商场只卖A&B可知B=(1-p%)Q;

代入等式A+1.25B=TR,可得p%Q+1.25*(1-p%)Q=(p/r)Q,即r=p/[p%+1.25*(1-p%)]=400p/(500-p)。

[该题应注意,由于题干一直在讲r percent & p percent,用等式表达时应注意表示成r% & p%~~]

板凳
 楼主| 发表于 2008-2-8 19:29:00 | 只看该作者

谢谢!!!

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